5
$\begingroup$

After reading this very interesting post about the electric field and the electric potential of a point charge in 2D and 1D, I've understood that, for the $2D-$case, the following formulas hold: $$ \Phi_{\operatorname{2-d}}(r) = -\frac{\lambda}{2\pi \epsilon_0} \ln(r) $$

$$ \vec{E}_{\operatorname{2-d}}(r) = \frac{\lambda}{\epsilon_0} \left(\frac{\hat{r}}{2\pi r}\right) $$

Nevertheless, I haven't fully understood how the dimensional analysis and the numerical values of the quantities that come into play (namely, the electric charge $\lambda$ and the vacuum permittivity $\epsilon_0$) change due to the reduced dimensionality of the system.

In other words:

  1. Is the value of the vacuum permittivity still $8.85 \, 10^{-12}\, F/m$, as in the usual 3D world?
  2. Is the value of a single electric charge still $1.60 \, 10^{-19} C$, as in the 3D world?
$\endgroup$

2 Answers 2

3
$\begingroup$

How you shuffle the units in different dimensions is a matter of opinion. If you keep the units of $\epsilon_0$ fixed, then the elementary charge in each dimensionality has units $\operatorname{C}\operatorname{m}^{d-3}$. Interestingly, the units of force and energy are similarly not independent of dimension. If I place two infinite line charges parallel to each other I cannot talk intelligently about the total force acting between them; it, like the total charges, is infinite. I, therefore, have to work in terms of force per unit length (similarly for energy).

The reason it works this way is because mass has a similar dimension dependence to its considerations as charges do, $\operatorname{kg} m^{d-3}$. Thus, you also have to modify the definition of $\vec{F}=m\vec{a}$. Now, $\vec{a}$ is clearly dimension independent (it's not extensive).

That's why we usually keep the units of $\epsilon_0$ dimension independent.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for the great explanation. Can you please comment also on the numerical value of $\epsilon_0$ and of $e$ (the elementary electric charge)? Does their numerical value change in reduced dimensionality? $\endgroup$
    – AndreaPaco
    Apr 15, 2019 at 9:43
  • 1
    $\begingroup$ @AndreaPaco The numerical values of $\epsilon_0$ and $e$ are meaningless - they're just conventions that depend on how we define our units. The difference between one value of $\epsilon_0$ and another are no different than measuring in feet versus inches versus meters. $\endgroup$ Apr 15, 2019 at 9:45
0
$\begingroup$

It would be more physical and intuitive if you consider the electric field of an infinite line in 3D with a constant charge density $\lambda$. Hence the unit is C/m and vacuum permittivity does not change. So, it is not a point charge in 2D but a charge density of an infinite line in 3D. This is why the symbol $\lambda$ is used.
Physical problem is defined in 3D, but mathematically, it reduces to 2D since going along the infinite line does not change the answer.
Similarly, for an infinite plane, the charge density per unit area is denoted by $\sigma$ with the unit C/$m^2$.
Value of the vacuum permittivity or the unit of the electric field does not change at all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.